# Silly Question about “opposite groups”

Given a group G, we can define Gop to be a new group whose underlying set is the same as that of G but with the new multiplication g.h = hg, i.e. multiply as if you were in G but reverse the order.

When is a group isomorphic to its opposite group?

One sees the "opposite ring" of a non-commutative ring showing up in a crucial way in number theory in the study of Brauer groups, so this is not necessarily an artificial question. This is just a question that has been poking around my brain for a little while. . . it seems like a potentially neat little problem.

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Every group is isomorphic to its opposite group, the map $g \mapsto g^{-1}$ being an isomorphism.

In ring theory this is not the case, and the opposite ring is an important construction. For instance, if $A$ is a central simple algebra over a field (or an Azumaya algebra over a ring), then the classes of $A$ and $A^{\operatorname{op}}$ are inverse in the Brauer group.

I don't know of anything similarly interesting to say about opposite groups.

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of course! how silly of me! –  Joel Dodge Oct 26 '09 at 5:19
You know,it's little facts born of simple questions like this that make algebra endlessly fascinating to me. –  Andrew L Aug 1 '10 at 4:50